Consider the arithmetic sequence presented in the table below. n 4 37 an 45 606 Hint: an = a1 + d(n − 1), where a1 is the first term and d is the common difference. Part A: What is the first term, a1, of the sequence? Part B: What is the general term equation, an, for this sequence? Part C: What is the value of the 13th term of this sequence?

Filling in the values from the table, you have two equations in two unknowns.
  37 = a1 + d·3
  606 = a1 + d·44
These equations can be solved by subtracting the first from the second.
  606 -37 = d(44 -3)
  569/41 = d = 13 36/41

Part A:
Rearranging the first equation, we have
  a1 = 37 - 3·d
  a1 = 37 - 3·(13 36/41)
  a1 = -4 26/41

Part B:
The general term can be found by filling the numbers into the formula given in the problem statement.
  an = (-4 26/41) + (13 36/41)·(n -1)

Part C:
The 13th term is found using the formula of Part B with n=13.
  a13 = -4 26/41 + (13 36/41)·12
  a13 = 161 37/41

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These numbers are unexpected. We find they become relatively nice integers if we assume there is an error in the problem statement. If the first table entry is supposed to be (4, 32),* then the results are found by the same methods, but turn out somewhat different:

A: a1 = -10
B: an = -10 + 14(n -1)
C: a13 = 158

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* We note it would not be difficult to mistake a hand-written 2 for a 7. The transcription error probably occurred early in the process of developing the curriculum material, and no checking was done of the final published work. It happens too often.